The two hemispheres at the ends will make up one complete sphere
Let the radius of the sphere be r ft, and let the height of the cylindrical part be h
Volume = cylinder + sphere
πr^2 h + (4/3)πr^3 = 144π
3πr^2 h + 4πr^3 = 432π
3r^2 h + 4r^3 = 432
h = (432 - 4r^3)/(3r^2)
The cost of production is dependent on the surface area of material used
SA = 4π^2 + 2πrh
= 4πr^2 + 2π(432 - 4r^3)/(3r^2)
= 4πr^2 + 288π/r^2 - (8/3)πr
d(SA)/dr = 8πr - 576π/r^3 - 8π/3 = 0 for a min of SA
8r - 576/r^3 - 8/3 = 0
24r^4 -1728 - 8r^3 = 0
3r^4 - r^3 - 216 = 0
At this point I "cheated" a bit and ran it through Wolfram
for a real solution of r=3, r = appr-3 and 2 complex roots,
so the only usable answer is r = 3
the h = (432 - 108)/27 = 12
So the cylinder part should be 12 ft long and the radius of each of the semispheres should be 3 ft
remember the sphere surface is twice as expensive as the cylinder which has no compound curvature so is easy to roll.
c = 2(4 pi r^2) + 1(2 pi r h)
c= 8 pi r^2 + 2 pi r h
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